Identify major product A in the reaction below
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2,4,6-tribromoaniline
The –\( \ce{NH2} \) group of aniline donates its lone pair into the ring by resonance. This makes the ortho and para positions centres of high electron density, so –\( \ce{NH2} \) is a powerful activating, ortho/para directing group. The ring is so reactive that bromine water substitutes all three ortho and para positions at room temperature, without a catalyst.
So A is 2,4,6-tribromoaniline.
Why the other options are incorrect
Option (1) 3-bromoaniline — Incorrect. –\( \ce{NH2} \) directs to ortho and para positions, not to the meta position.
Option (2) 2-bromoaniline — Incorrect. In bromine water the reaction does not stop at one substitution; the very active ring is brominated at all three positions.
Option (3) 4-bromoaniline — Incorrect. To get the monobromo product the activation must first be reduced by acetylating –\( \ce{NH2} \) to –\( \ce{NHCOCH3} \); brominating acetanilide and then hydrolysing gives 4-bromoaniline.
NCERT (2026-27 edition):
Bromination
